On the creation of wave packets with particular properties in quantum field theory

Question

At the start of chapter 5 of Mark Srednicki's lecture notes on quantum field theory we define an operator that creates a particle that is "localised in momentum space near $\mathbf {k_1}$, and localised in position space near the origin":

$$a_1^\dagger\equiv\int d^3k\text{ }f_1(\mathbf k)a^\dagger(\mathbf k) \tag{5.6},$$

in which:

$$f_1(\mathbf k)\propto \exp[-(\mathbf k-\mathbf {k_1})^2/4\sigma^2] \tag{5.7}.$$

I do not follow how this necessarily creates a wave packet with the required properties. I see that a related question has already been asked on the site, but the answer doesn't address what I'm asking. I understand that we want the particle to be localised in position space so that its asymptotic behaviour allows us to consider its interactions perturbatively, but what specifically about the above construction makes these particles "localised in momentum/position space"?

Are you cool with $\langle {\mathbf k}| a_1^\dagger |0\rangle$? — Cosmas Zachos, Nov 19 '20 at 20:24
@CosmasZachos I think $\langle \mathbf k|a_1^\dagger|0\rangle=(2\pi)^3\omega\delta^3(\mathbf k-\mathbf k')$? Which I realise now is written on the same page, but I do understand it. — Charlie, Nov 19 '20 at 20:29

score 5 · Accepted Answer · answered Nov 19 '20 at 22:49

5

This is not an essentially QFT question, but rather a question of quantum mechanics.

The point is simply

that Gaussians minimize the position/momentum uncertainty (see the answer to this question)
that Gaussians go to Gaussians under Fourier transform.

Fourier transform (5.7) to find a Gaussian (in position space) which should be localised (peaking at) not too far away from zero.

answered Nov 19 '20 at 22:49

1

This helps a lot, I hadn't thought about those two points, thanks! – Charlie Nov 19 '20 at 23:28

On the creation of wave packets with particular properties in quantum field theory

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